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How To Calculate Solar Panels Needed
Sizing a solar array is a division problem. Work out how much energy one panel produces where you live, then divide your daily consumption by that number.
Quick Answer
panels = ceil(daily kWh / (panel kW x peak sun hours x (1 - losses)))
- daily kWh
- Your average electricity use per day
- peak sun hours
- The equivalent hours of full strength sun per day
- panel kW
- The rated output of one panel, watts divided by 1000
- losses
- Everything lost to wiring, heat, dirt and the inverter
A 400 watt panel in a place with 4.5 peak sun hours and 14 percent system losses produces 1.548 kWh a day. A home using 30 kWh a day therefore needs 20 panels, which is an 8 kW system covering about 430 square feet.
What Is Solar Panels Needed?
Every solar sizing starts from consumption, not from roof size. The question is how much energy you use in a day, and how much one panel will give back where you live.
Annual consumption divided by 365 gives the daily figure. A home on 10950 kWh a year uses 30 kWh a day, and that single number drives the whole calculation.
Peak sun hours is not the length of the day. It is the equivalent number of hours at full strength, so a long cloudy day might contribute two peak sun hours while a short clear one contributes six.
Most of the United States sits between three and six peak sun hours a year round, and the figure varies by more than a factor of two between the north and the south west.
A panel's rating is measured under laboratory conditions. Four hundred watts means 400 watts at a cell temperature of 25 degrees and full perpendicular sun, which is not what a roof gives you.
Real output is lower, and that gap is the system loss. Inverter conversion, cable resistance, dust, shading and the fact that hot panels produce less all come off the top, typically around 14 percent.
So the daily output of one panel is its kilowatt rating times the peak sun hours times one minus the losses. For a 400 watt panel at 4.5 hours and 14 percent loss that is 1.548 kWh a day.
Divide your daily use by that and round up. Thirty divided by 1.548 is 19.38, so 20 panels, because you cannot buy a fraction of one and because the array should cover the bad months too.
The system size follows directly: 20 panels of 400 watts is 8 kilowatts, which is the number installers quote and the number rebates are usually based on.
Roof area is the next constraint. A 400 watt panel covers roughly 21.5 square feet, so 20 of them want about 430 square feet, before setbacks and walkways are taken into account.
Roof orientation and tilt move the peak sun hours figure. A south facing roof at the right tilt gets the full value, while an east or west roof typically delivers ten to twenty percent less.
Coverage above 100 percent is normal and intentional. Because losses are already inside the calculation, a system that reports 103.2 percent of annual usage is sized to cover a typical year, not a perfect one.
Self consumption matters more than total production. A grid tied home exports its midday surplus, so what you save depends on when you use power as much as how much you generate.
Batteries change the arithmetic. Storage lets the surplus be used after dark, which raises self consumption but does not by itself reduce the number of panels needed.
Size for the loads you expect, not only the ones you have. An electric car adds roughly 3000 kWh a year and a heat pump can add more, so building in headroom now is far cheaper than adding panels later.
Formula
daily kWh = (watts / 1000) x peak sun hours x (1 - losses)
Formula page →The rated output, scaled by the local sun and reduced by real world losses.
| Symbol | Meaning | Unit | Notes |
|---|---|---|---|
| E | Daily output | kWh | The energy one panel makes per day. |
| W | Panel rating | watts | The nameplate watts of the panel. |
| H | Peak sun hours | hours | Equivalent hours of full sun per day. |
| losses | System losses | decimal | The fraction lost to heat, wiring, dirt and conversion. |
panels = ceil(daily usage / daily output per panel)
Formula page →Daily consumption divided by what one panel delivers, rounded up.
| Symbol | Meaning | Unit | Notes |
|---|---|---|---|
| N | Panels | panels | The number of panels to install. |
| U | Daily usage | kWh | The average energy used per day. |
| E | Panel output | kWh | The daily energy from one panel. |
roof area = panels x panel area, coverage = annual output / annual usage
Formula page →Turns the panel count into a footprint and a percentage of your usage.
| Symbol | Meaning | Unit | Notes |
|---|---|---|---|
| A | Roof area | square feet | The footprint the array occupies. |
| a | Panel area | square feet | The area of one panel. |
| C | Coverage | decimal | Annual production as a share of annual usage. |
How To Calculate Solar Panels Needed
- 1
Find your daily electricity use
Read the annual total off your bill and divide by 365. A home using 10950 kWh a year uses 30 kWh a day, which is the number the array has to cover.
- 2
Work out what one panel produces
A 400 watt panel is 0.4 kilowatts. At 4.5 peak sun hours with 14 percent system losses it delivers 0.4 times 4.5 times 0.86, which is 1.548 kWh a day.
- 3
Divide and round up
Thirty divided by 1.548 is 19.38 panels, so 20 go on the roof. Rounding up matters because a short array leaves you buying grid power at the worst tariff.
- 4
Check the roof and the coverage
Twenty panels at 21.5 square feet each need about 430 square feet, and the array produces 11300.4 kWh a year, which is 103.2 percent of the 10950 kWh the home uses.
Examples
Example 1: A home using 30 kWh a day with 400 watt panels
- Daily electricity use
- 30
- Peak sun hours
- 4.5
- Panel rating
- 400
- System losses
- 14
- Panel area
- 21.5
| Step | Calculation | Result |
|---|---|---|
| Panel rating in kilowatts | 400 / 1000 | 0.4 |
| Daily output of one panel | 0.4 x 4.5 x 0.86 | 1.548 |
| Panels before rounding | 30 / 1.548 | 19.3798 |
| Panels to install | ceil(19.3798) | 20 |
Result: One 400 watt panel delivers 1.548 kWh a day at this location, so 30 kWh of daily use needs 20 panels. That is an 8 kilowatt system producing 11300.4 kWh a year, or 103.2 percent of usage.
Example 2: A smaller home using 12 kWh a day with 350 watt panels
- Daily electricity use
- 12
- Peak sun hours
- 5
- Panel rating
- 350
- System losses
- 14
- Panel area
- 18.5
| Step | Calculation | Result |
|---|---|---|
| Panel rating in kilowatts | 350 / 1000 | 0.35 |
| Daily output of one panel | 0.35 x 5 x 0.86 | 1.505 |
| Panels before rounding | 12 / 1.505 | 7.9734 |
| Panels to install | ceil(7.9734) | 8 |
Result: Each 350 watt panel makes 1.505 kWh a day in a five peak sun hour location, so 12 kWh of daily use needs 8 panels. That is a 2.8 kilowatt array covering about 148 square feet of roof.
Calculator
Panels needed
20
- System size
- 8
- Daily output per panel
- 1.548
- Annual output
- 11,300.4
- Share of usage covered
- 103.20%
- Roof area needed
- 430
Values update as you type. This calculator covers the single scenario its formula assumes — see Common Mistakes for what it leaves out.
Prefer a full-width tool? Open the Solar Panels Needed calculator page.
Common Mistakes
Treating the panel rating as real output
A 400 watt panel rarely delivers 400 watts on a roof. Heat alone costs several percent, and the rating assumes a cool panel in laboratory light.
Ignoring system losses
Inverter conversion, cable runs, dust and shading typically take 14 percent. Leaving them out under sizes the array by roughly one panel in seven.
Confusing day length with peak sun hours
A fourteen hour summer day is not fourteen peak sun hours. The figure is the equivalent hours at full strength, which is four to six in most places.
Forgetting the roof area and the setbacks
Panels need space, and fire codes require gaps at the ridge and edges. A roof that is 430 square feet on paper may only accept 350 square feet of array.
Sizing only for today's consumption
An electric car or a heat pump can add thousands of kilowatt hours a year. Adding panels later costs more per panel than building the headroom in now.
Using an annual average without checking the seasons
Production swings widely between summer and winter. An array that matches the annual average exactly will still import power through the short days.
FAQ
How many solar panels do I need?
Divide your daily kilowatt hours by what one panel makes per day. A home on 30 kWh a day with 400 watt panels and 4.5 peak sun hours needs 20 panels.
What are peak sun hours?
The equivalent number of hours per day at full strength sunlight. Four to six covers most of the United States, and the figure already accounts for cloud and low sun.
How much roof area does an array need?
Multiply the panel count by the area of one panel. Twenty panels at 21.5 square feet need about 430 square feet, plus the setbacks your fire code requires.
Why does my system cover more than 100 percent of my usage?
Because losses are already inside the panel output figure, a result a little above 100 percent is a typical year, not a perfect one. It also leaves room for a hot summer or a snowy week.
Do I need batteries?
Not for the panel count. Batteries raise how much of your own generation you actually use, which changes the savings rather than the size of the array.
What losses should I assume?
About 14 percent is a common planning figure, covering the inverter, cable resistance, dust, shading and the drop in output as panels heat up.
References
- [1]Wikipedia, Solar panel — https://en.wikipedia.org/wiki/Solar_panel
- [2]Wikipedia, Photovoltaic system — https://en.wikipedia.org/wiki/Photovoltaic_system
- [3]Wikipedia, Insolation — https://en.wikipedia.org/wiki/Insolation